Search Results

Search found 21184 results on 848 pages for 'oracle oracle vm templates'.

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

Page 391/848 | < Previous Page | 387 388 389 390 391 392 393 394 395 396 397 398  | Next Page >

  • Can the template parameters of a constructor be explicitly specified?

    - by sth
    A constructor of a class can be a template function. At the point where such a constructor is called, the compiler usually looks at the arguments given to the constructor and determines the used template parameters from them. Is there also some syntax to specify the template parameters explicitly? A contrived example: struct A { template<typename T> A() {} }; Is there a way to instantiate this class? What is the syntax to explicitly specify the constructor's template parameters? My use case would be a problem were the compiler doesn't seem to find the correct templated constructor. Explicitly specifying the template parameters would probably generate more useful error messages or even resolve the problem.

    Read the article

  • Problems Expanding an Array in C++

    - by dxq
    I'm writing a simulation for class, and part of it involves the reproduction of organisms. My organisms are kept in an array, and I need to increase the size of the array when they reproduce. Because I have multiple classes for multiple organisms, I used a template: template <class orgType> void expandarray(orgType* oldarray, int& numitems, int reproductioncount) { orgType *newarray = new orgType[numitems+reproductioncount]; for (int i=0; i<numitems; i++) { newarray[i] = oldarray[i]; } numitems += reproductioncount; delete[] oldarray; oldarray = newarray; newarray = NULL; } However, this template seems to be somehow corrupting my data. I can run the program fine without reproduction (commenting out the calls to expandarray), but calling this function causes my program to crash. The program does not crash DURING the expandarray function, but crashes on access violation later on. I've written functions to expand an array hundreds of times, and I have no idea what I screwed up this time. Is there something blatantly wrong in my function? Does it look right to you? EDIT: Thanks for everyone's help. I can't believe I missed something so obvious. In response to using std::vector: we haven't discussed it in class yet, and as silly as it seems, I need to write code using the methods we've been taught.

    Read the article

  • C++ Iterators and inheritance

    - by jomnis
    Have a quick question about what would be the best way to implement iterators in the following: Say I have a templated base class 'List' and two subclasses "ListImpl1" and "ListImpl2". The basic requirement of the base class is to be iterable i.e. I can do: for(List<T>::iterator it = list->begin(); it != list->end(); it++){ ... } I also want to allow iterator addition e.g.: for(List<T>::iterator it = list->begin()+5; it != list->end(); it++){ ... } So the problem is that the implementation of the iterator for ListImpl1 will be different to that for ListImpl2. I got around this by using a wrapper ListIterator containing a pointer to a ListIteratorImpl with subclasses ListIteratorImpl2 and ListIteratorImpl2, but it's all getting pretty messy, especially when you need to implement operator+ in the ListIterator. Any thoughts on a better design to get around these issues?

    Read the article

  • Specializing function template for both std::string and char*

    - by sad_man
    As the title says I want to specialize a function template for both string and char pointer, so far I did this but I can not figure out passing the string parameters by reference. #include <iostream> #include <string> template<typename T> void xxx(T param) { std::cout << "General : "<< sizeof(T) << std::endl; } template<> void xxx<char*>(char* param) { std::cout << "Char ptr: "<< strlen(param) << std::endl; } template<> void xxx<const char* >(const char* param) { std::cout << "Const Char ptr : "<< strlen(param)<< std::endl; } template<> void xxx<const std::string & >(const std::string & param) { std::cout << "Const String : "<< param.size()<< std::endl; } template<> void xxx<std::string >(std::string param) { std::cout << "String : "<< param.size()<< std::endl; } int main() { xxx("word"); std::string aword("word"); xxx(aword); std::string const cword("const word"); xxx(cword); } Also template<> void xxx<const std::string & >(const std::string & param) thing just does not working. If I rearranged the opriginal template to accept parameters as T& then the char * is required to be char * & which is not good for static text in code. Please help !

    Read the article

  • Declare variables that depend on unknown type in template functions.

    - by rem
    Suppose I'm writing a template function foo that has type parameter T. It gets an object of type T that must have method bar(). And inside foo I want to create a vector of objects of type returned by bar. In GNU C++ I can write something like that: template<typename T> void foo(T x) { std::vector<__typeof(x.bar())> v; v.push_back(x.bar()); v.push_back(x.bar()); v.push_back(x.bar()); std::cout << v.size() << std::endl; } How to do the same thing in Microsoft Visual C++? Is there some way to write this code that works in both GNU C++ and Visual C++?

    Read the article

  • Is it possible to supply template parameters when calling operator()?

    - by Paul
    I'd like to use a template operator() but am not sure if it's possible. Here is a simple test case that won't compile. Is there something wrong with my syntax, or is this simply not possible? struct A { template<typename T> void f() { } template<typename T> void operator()() { } }; int main() { A a; a.f<int>(); // This compiles. a.operator()<int>(); // This compiles. a<int>(); // This won't compile. return 0; }

    Read the article

  • template function error..

    - by sil3nt
    Hi there, I have function which takes in an parameter of a class called "Triple", and am returning the averge of 3 values of type float. template <typename ElemT> float average(Triple ElemT<float> &arg){ float pos1 = arg.getElem(1); float pos2 = arg.getElem(2); float pos3 = arg.getElem(3); return ( (pos1+pos2+po3) /3 ); } when i try compiling this i get q2b.cpp:32: error: template declaration of `float average' q2b.cpp:32: error: missing template arguments before "ElemT" not quite sure what this means.

    Read the article

  • C++, what does this syntax mean?

    - by aaa
    i found this in this file: http://www.boost.org/doc/libs/1_43_0/boost/spirit/home/phoenix/core/actor.hpp What does this syntax means? struct actor ... { ... template <typename T0, typename T1> typename result<actor(T0&,T1&)>::type // this line thank you

    Read the article

  • How to save links with tags and parameters in TextField

    - by xRobot
    I have this simple Post model: class Post(models.Model): title = models.CharField(_('title'), max_length=60, blank=True, null=True) body = models.TextField(_('body')) blog = models.ForeignKey(Blog, related_name="posts") user = models.ForeignKey(User) I want that when I insert in the form the links, then these links are saved in the body from this form: http://www.example.com or www.example.com to this form ( with tag and rel="nofollow" parameter ): <a href="http://www.example.com" rel="nofollow">www.example.com</a> How can I do this ? Thanks ^_^

    Read the article

  • C++: How to require that one template type is derived from the other

    - by Will
    In a comparison operator: template<class R1, class R2> bool operator==(Manager<R1> m1, Manager<R2> m2) { return m1.internal_field == m2.internal_field; } Is there any way I could enforce that R1 and R2 must have a supertype or subtype relation? That is, I'd like to allow either R1 to be derived from R2, or R2 to be derived from R1, but disallow the comparison if R1 and R2 are unrelated types.

    Read the article

  • Class templating std::set key types

    - by TomFLuff
    I have a class to evaluate set algebra but wish to template it. At the minute it looks a bit like this set.h: template<typename T> class SetEvaluation { public: SetEvaluation<T>(); std::set<T> evaluate(std::string in_expression); } set.cpp template<typename T> std::set<T> SetEvaluation<T>::evaluate(std::string expression) { std::set<T> result; etc etc... } But i'm getting undefined reference errors when compiling. Is it possible to declare the return type as std::set<T> and then pass std::string as the class template param. There are no errors in the class but only when I try to instantiate SetEvaluation<std::string> Can anyone shed light on this problem? thanks

    Read the article

  • Usage of CRTP in a call chain

    - by fhw72
    In my widget library I'd like to implement some kind of call chain to initialize a user supplied VIEW class which might(!) be derived from another class which adds some additional functionality like this: #include <iostream> template<typename VIEW> struct App { VIEW view; void init() {view.initialize(); } }; template<typename DERIVED> struct SpecializedView { void initialize() { std::cout << "SpecializedView" << std::endl; static_cast<DERIVED*>(this)->initialize(); } }; struct UserView : SpecializedView<UserView> { void initialize() {std::cout << "UserView" << std::endl; } }; int _tmain(int argc, _TCHAR* argv[]) { // Cannot be altered to: App<SpecializedView<UserView> > app; App<UserView> app; app.init(); return 0; } Is it possible to achieve some kind of call chain (if the user supplied VIEW class is derived from "SpecializedView") such that the output will be: console output: SpecializedView UserView Of course it would be easy to instantiate variable app with the type derived from but this code is hidden in the library and should not be alterable. In other words: The library code should only get the user derived type as parameter.

    Read the article

  • Templated derived class in CRTP (Curiously Recurring Template Pattern)

    - by Butterwaffle
    Hi, I have a use of the CRTP that doesn't compile with g++ 4.2.1, perhaps because the derived class is itself a template? Does anyone know why this doesn't work or, better yet, how to make it work? Sample code and the compiler error are below. Source: foo.C #include <iostream> using namespace std; template<typename X, typename D> struct foo; template<typename X> struct bar : foo<X,bar<X> > { X evaluate() { return static_cast<X>( 5.3 ); } }; template<typename X> struct baz : foo<X,baz<X> > { X evaluate() { return static_cast<X>( "elk" ); } }; template<typename X, typename D> struct foo : D { X operator() () { return static_cast<D*>(this)->evaluate(); } }; template<typename X, typename D> void print_foo( foo<X,D> xyzzx ) { cout << "Foo is " << xyzzx() << "\n"; } int main() { bar<double> br; baz<const char*> bz; print_foo( br ); print_foo( bz ); return 0; } Compiler errors foo.C: In instantiation of ‘foo<double, bar<double> >’: foo.C:8: instantiated from ‘bar<double>’ foo.C:30: instantiated from here foo.C:18: error: invalid use of incomplete type ‘struct bar<double>’ foo.C:8: error: declaration of ‘struct bar<double>’ foo.C: In instantiation of ‘foo<const char*, baz<const char*> >’: foo.C:13: instantiated from ‘baz<const char*>’ foo.C:31: instantiated from here foo.C:18: error: invalid use of incomplete type ‘struct baz<const char*>’ foo.C:13: error: declaration of ‘struct baz<const char*>’

    Read the article

  • templated class : accessing derived normal-class methods

    - by user1019129
    I have something like this : class Container1 { public: method1() { ... } } class Container2 { public: method1() { ... } } template<class C = Container1> class X : public C { public: using C::method1(); ..... X(string& str) : C(str) {}; X& other_method() { method1(); ...; } } My question is why I have to use "using C::method1()", to be able to access the method.. Most of answers I found is for the case where templated-class inhering templated-class. Normally they mention using "this-", but this does not seem to work in this case. Can I do something else shorter... Also I'm suspecting the other error I'm getting is related to the same problem : no match call for (X<Container1>) (<std::string&>)

    Read the article

  • Django template CSS/IMG is "off" in the URL

    - by erimar77
    I have /path/to/my/theme/static/css/frontend.css which is called by base.html <link rel="stylesheet" type="text/css" href="{{ STATIC_URL }}css/frontend.css" media="all" /> In which I've got a background for the header: #header-wrapper min-width: 960px; height: 150px; background: transparent url(img/header-bg.png) repeat-x center bottom; } The file is /path/to/my/theme/static/img I've run manage.py collectstatic to gather the files and almost everything looks correct except the link generated looks like: http://example.com/static/css/img/header-bg.png In which the image does not show, because the correct URL is: http://example.com/static/img/header-bg.png Where am I going wrong??

    Read the article

  • Django: Is it possible to attach media files (css, javascript etc) to a View-class?

    - by mrmclovin
    I can't fins any information on how to define css or javascript files in a view like: class MyView(View): .... class Media: css = { 'all' : 'mystyle.css' } If you have a form you can do like: class MyForm(ModelForm): .... class Media: css = { 'all' : 'mystyle.css' } And then in the template you can print the files like; {{ form.media.css }} I like that Syntax very much and I like to keep the View-specific css files in the app-directory. Does anyone know if it's possible? Thanks!

    Read the article

  • Why can't I create a templated sublcass of System::Collections::Generic::IEnumerable<T>?

    - by fiirhok
    I want to create a generic IEnumerable implementation, to make it easier to wrap some native C++ classes. When I try to create the implementation using a template parameter as the parameter to IEnumerable, I get an error. Here's a simple version of what I came up with that demonstrates my problem: ref class A {}; template<class B> ref class Test : public System::Collections::Generic::IEnumerable<B^> // error C3225... {}; void test() { Test<A> ^a = gcnew Test<A>(); } On the indicated line, I get this error: error C3225: generic type argument for 'T' cannot be 'B ^', it must be a value type or a handle to a reference type If I use a different parent class, I don't see the problem: template<class P> ref class Parent {}; ref class A {}; template<class B> ref class Test : public Parent<B^> // no problem here {}; void test() { Test<A> ^a = gcnew Test<A>(); } I can work around it by adding another template parameter to the implementation type: ref class A {}; template<class B, class Enumerable> ref class Test : public Enumerable {}; void test() { using namespace System::Collections::Generic; Test<A, IEnumerable<A^>> ^a = gcnew Test<A, IEnumerable<A^>>(); } But this seems messy to me. Also, I'd just like to understand what's going on here - why doesn't the first way work?

    Read the article

  • C++ Template Question

    - by user323422
    see following code and please clear doubts1. as ABC is template why it not showing error when we put defination of ABC class member function in test.cpp 2.if i put test.cpp code in test.h , then it working fine // test.h template <typename T> class ABC { public: void foo( T& ); void bar( T& ); }; // test.cpp template <typename T> void ABC<T>::foo( T& ) {} // definition template <typename T> void ABC<T>::bar( T& ) {} // definition template void ABC<char>::foo( char & ); // 1 // main.cpp #include "test.h" int main() { ABC<char> a; a.foo(); // working a.bar(); // link error }

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

< Previous Page | 387 388 389 390 391 392 393 394 395 396 397 398  | Next Page >